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设y=根号x分之x²-x+根号x-1,求导数与微分?
人气:359 ℃ 时间:2019-10-19 16:35:38
解答
y=(x²-x)/√x+√(x-1)
=x^(3/2)-√x+√(x-1)
所以
y'=3/2*√x-1/2√x+1/2√(x-1)
=(3x-1)/(2√x)+1/2√(x-1)
dy/dx=y'
所以
dy=[(3x-1)/(2√x)+1/2√(x-1)]dx
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